# Lebesgue's Dominated Convergence Theorem

I have found an example: What is $$\displaystyle\lim_{n\rightarrow\infty}\int_{0}^{\infty}\frac{1}{n}\chi_{[0,n)}\,dx$$ ?

I mean, I know it is $$1$$, because for every $$n$$, the value of the integral is $$1$$, and the limit of the constant $$1$$ function is also $$1$$. However, I don't understand why I can't use the Dominated Convergence Theorem!? If I change the $$\lim$$ and $$\int$$ operators, I get $$0$$, since $$\lim_{n\rightarrow\infty}\frac{1}{n}\chi_{[0,n)}=0$$, and if I integrate $$0$$, it is going to be $$0$$.

Therefore, I miss something from the Dominated Convergence Theorem, or I don't understand it clearly. Why can't I change the the order of operators?

• You need a dominating function which is integrable. – Thomas Shelby Aug 7 '19 at 8:47

DCT requires the condition that the integrand is dominated by an integarble function. In this case if $$\frac 1 n I_{[n, \infty)} \leq g$$ then $$g \geq \frac 1 n$$ on the interval $$[n,\infty)$$ for all $$n$$ and, in particular for $$n=1$$, so $$\int_0^{\infty} g =\infty$$. Thus there is no dominating integrable function,.